molL1HAc溶液稀释到什么体积才能使解离度增加一倍
第四章习题
?????
,11.将300mL0.20 mol,LHAc溶液稀释到什么体积才能使解离度增加一倍。
,1,,0.20molL300mL解:设稀释到体积为V ,稀释后 ,c V
222,0.20,300,(2),c ,0.20 ,由 aK, 得: ,1,,1,,V,(1,2,)
,5 ,1,,,,因为K=1.74,10c = 0.2 mol,L cK> 20K c/K>500 a aaa waa
故由 1,2, =1,, 得 V =[300,4/1]mL =1200mL
,,,此时仍有 cK>20K c/K>500 。 aawaa,1,2.求算 0.20mol,LNHHO的c(OH)及解离度。 32,5 ,,,,解:K(NH?HO)=1.74,10 由于cK>20K, c/K>500 b32bbwbb
,由 c(OH),c,K bb
,,5,1,3,1得 mol,L=1.9,10 mol,L c(OH),0.20,1.74,10
,,3(OH)c,31.9,10 ,,,,9.5,10,0.95%0.20cb
,13.奶油腐败后的分解产物之一为丁酸(CHCOOH),有恶臭。今有0.40L含0.20 mol,L37
,丁酸的溶液, pH为2.50, 求丁酸的K。 a+,2.5,1解:pH=2.50 c(H) =10 mol,L
,2.5,2, =10/0.20 = 1.6,10
,2220.20,(1.6,10),,5c,K= ,,5.2,10a,21,,1,1.6,10,1,4. What is the pH of a 0.025mol,L solution of ammonium acetate at 25?? pK of acetic a,acid at 25? is 4.76, pK of the ammonium ion at 25? is 9.25, pK is 14.00. aw
+,4.76,9.24,7.00解: c(H)= KK,10,10,1012aa
+pH= ,logc(H) = 7.00
,,5.已知下列各种弱酸的K值,求它们的共轭碱的K值,并比较各种碱的相对强弱。 ab,10,4,,(1)HCN K =6.2×10; (2)HCOOH K =1.8×10; aa,5,10,,(3)CHCOOH(苯甲酸) K =6.2×10; (4) CHOH (苯酚) K =1.1×10; 65a65a,10,2,5,,,(5)HAsOK =6.0×10; (6) HCO K=5.9,10;K=6.4,10; 2 a224a1a2,10,10 ,5 解: (1)HCN K= 6.2,10 K =K/6.2,10=1.6,10 abw,4,4,11(2)HCOOH K= 1.8,10 K=K /1.8,10 =5.6,10 abw,5 ,5 ,10(3)CHCOOH K= 6.2,10 K=K /6.2,10=1.61×10 65abw,10,10,5 (4)CHOH K=1.1,10 K =K /1.1,10 =9.1,10 65abw,10,10,5 (5)HAsO K=6.0,10 K =K /6.0,10 =1.7,10 2abw,2 ,2 ,13(6)HCO K=5.9,10 K =K /5.9,10=1.7,10 224a1b2w,5 ,5 ,10 K=6.4,10 K=K /6.4,10=1.5×10 a2b1w
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无机及分析化学学习指导
,,,, ,2,, 碱性强弱:CHO > AsO > CN > CHCOO>CO > HCOO> HCO652652424
6.用质子理论判断下列物质哪些是酸?并写出它的共轭碱。哪些是碱?也写出它的共轭
酸。其中哪些既是酸又是碱?
-----2 HPO,CO,NH,NO,HO,HSO,HS,HCl 2433324
解:
酸 共轭碱 碱 共轭酸 既是酸又是碱
---2,HPO HPO PO HPO HHPO 342424244+,NH NH NH NHNH 33432+,HO O HO HO HOH 2232---2,HSOSO HHSO HSOSO 244444---2,HSS HHS HSS 2-,HCl Cl HNO NO 33
2,, CO HCO 33
7.写出下列化合物水溶液的PBE:
(1) HPO (2) NaHPO (3) NaS (4)NHHPO 34242424
(5) NaCO (6) NHAc (7) HCl+HAc (8)NaOH+NH 22443解:
+, ,2,3,(1) HPO:c( H) = c(HPO ) + 2c( HPO) + 3c (PO) + c(OH) 342444+,,3,(2) NaHPO: c(H) + c(HPO ) + 2c(HPO) = c(PO) + c(OH) 2424344,+,(3) NaS: c(OH) = c(H) + c(HS) + 2c(HS ) 22+ ,2,3,(4)NHHPO: c(H) + c(HPO) = c(NH) + c(HPO)+ 2c(PO) + c(OH) 42434344,+,(5)NaCO: c(OH) = c(H) + c(HCO) + 2c(HCO) 22424224+,(6)NHA: c(HAc) + c(H) = c(NH) + c(OH) 4C3+,,,(7)HCl + HAc: c(H) = c(Ac) + c(OH) + c(Cl )
+,+(8)NaOH + NH: c(NH) + c(H) = c(OH) – c(NaOH) 34
8.某药厂生产光辉霉素过程中,取含NaOH的发酵液45L (pH=9.0),欲调节酸度到pH=3.0,
,1问需加入6.0 mol,L HCl溶液多少毫升?
,,5,1,5解: pH = 9.0 pOH = 14.0 – 9.0 = 5.0 c(OH) =1.0, 10 mol,L n(NaOH)= 45,10 mol
,53,2 设加入V mLHCl 以中和NaOH V= [45,10/6.0]10mL = 7.5,10mL 11 +,3,1 设加入xmLHCl使溶液pH =3.0 c(H) =1,10 mol,L ,3,5 ,3,36.0, x10/(45+7.5,10+ x10 ) = 1,10 x = 7.5mL
,2 共需加入HCl:7.5mL + 7.5,10mL = 7.6mL
,2,1,9.HSO第一级可以认为完全电离,第二级K =1.2×10,,计算0.40 mol,L HSO溶24a224
液中每种离子的平衡浓度。
+,2,解: HSO H + SO 44,1 起始浓度/mol,L0.40 0.40 0
,1 平衡浓度/mol,L 0.40,x 0.40 +x x
,2,1 1.2,10 = x(0.40 + x)/(0.40 , x) x = 0.011 mol,L
+,1 c(H) = 0.40 + 0.011 = 0.41 mol,L pH = ,lg0.41 = 0.39
,1 ,1,2,c(HSO) = 0.40 , 0.011 = 0.39 mol,Lc(SO) = 0.011 mol,L 44
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第四章习题
,1,110.某一元酸与36.12mL 0.100 mol,L NaOH溶液中和后,再加入18.06mL 0.100 mol,L
HCl溶液,测得pH值为4.92。计算该弱酸的解离常数。
,1,3解:36.12mL0.100 mol,LNaOH与该酸中和后, 得其共轭碱n=3.612,10mol; b,1,3加入18.06mL0.100mol,LHCl后生成该酸n=1.806,10mol; a,3,3剩余共轭碱n=(3.612,1.806),10mol = 1.806,10mol b,4.92,5,,,pH = pK , lgc/c= pK = 4.92 K = 10 = 1.2,10 aabaa,6,111.求1.0×10 mol,LHCN溶液的pH值。(提示:此处不能忽略水的解离)
,10,,,,解:K(HCN)= 6.2,10 c,K<20K c/K,500 aaaw aa
,,6,10,14,7,1,, c(H),c,K,K,1.0,10,6.2,10,1.0,10,1.0,10mol,L aaw
pH = 7.0
,1,12.计算浓度为0.12mol,L 的下列物质水溶液的pH值(括号内为pK值): a(1) 苯酚(9.89); (2) 丙烯酸(4.25) (3) 氯化丁基胺( CHNHCl) (9.39); (4) 吡啶的硝酸盐(CHNHNO)(5.25) 493553
+,9.89,6 ,,解:(1) pK = 9.89 c( H) = pH = 5.41 c,K,0.12,10,3.9,10aaa
+,4.25,3 ,,(2) pK = 4.25 c( H) = pH = 2.59 c,K,0.12,10,2.6,10aaa
+,9.39,6,,(3) pK = 9.39 c( H) = pH = 5.15 c,K,0.12,10,7.0,10aaa
+,5.25,4, ,(4)pK = 5.25 c( H) = pH = 3.09 c,K,0.12,10,8.2,10aaa
--,8,,13.HPO的K= 6.2×10,则其共轭碱的K是多少?如果在溶液中c(HPO)和其共24a2 b24轭碱的浓度相等时,溶液的pH将是多少?
,14,8 ,7,,,解: K = K/K= 1.0,10/6.2,10=1.6,10 bwa ,8,,pH = pK , lgc/c= pK= ,lg(6.2,10) = 7.20 aab a
14.0.20mol的NaOH和0.20molNHNO溶于足量水中并使溶液最后体积为1.0 L,问此时43
溶液pH为多少。
,1,5,解:平衡后为0.20 mol,L的NH?HO溶液 K=1.74,10 32b,,,cK>20K c/K> 500 bb wbb
,,5,3,1c(OH) =mol,L c,K,0.20,1.74,10,1.87,10bb
pOH = 2.73 pH = 14.00 , 2.73 = 11.27
,115.欲配制250mL pH=5.0的缓冲溶液,问在125mL1.0 mol,LNaAc溶液中应加多少6.0 ,1mol,L的HAc和多少水?
,5解: pH = pK , lgc/c 5.0 = ,lg(1.74,10) , lgc/c aabab,1,1 c/c = 0.575 c =1.0 mol,L,125/250 = 0.50 mol,L abb,1,1 c = 0.50 mol,L,0.575 = 0.29 mol,L a,1,1 V,6.0mol,L = 250mL ,0.29mol,L V = 12mL
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无机及分析化学学习指导
,1即要加入12mL 6.0 mol,L HAc 及 250 mL ,125 mL ,12 mL =113mL水。
16.今有三种酸(CH)AsOH, ClCHCOOH,CHCOOH,它们的
解离常数分别为32223,7,5,56.4×10, 1.4×10 , 1.76×10。试问:
(1)欲配制 pH= 6.50缓冲溶液,用哪种酸最好?
(2)需要多少克这种酸和多少克NaOH以配制1.00L缓冲溶液,其中酸和它的共轭碱的
,1总浓度等于1.00mol,L?
解:(1)(CH)AsOH的pK = 6.19;ClCHCOOH的pK = 4.85;CHCOOH的pK = 4.76; 322a2a3a
配pH = 6.50的缓冲溶液选(CH)AsOH最好,其pK与pH值最为接近。 322a,1(2)pH = pK , lgc/c 6.50 = 6.19 , lg[c/(1.00,c)] c = 0.329 mol,L aabaaa,1,1,1c = 1.00,c = 1.00 mol,L ,0.329 mol,L= 0.671 mol,L ba,1,1应加NaOH: m(NaOH)= 1.00L ,0.671 mol,L,40.01g,moL=26.8g
,1 需(CH)AsOH:m((CH)AsOH) =1.00L ,138 g,moL=138g 322322,117.现有一份HCl溶液,其浓度为0.20 mol,L。
(1)欲改变其酸度到pH= 4.0应加入HAc还是NaAc?为什么?
,1(2)如果向这个溶液中加入等体积的2.0 mol,LNaAc溶液,溶液的pH是多少?
,1(3)如果向这个溶液中加入等体积的2.0 mol,LHAc溶液,溶液的pH是多少?
,1(4)如果向这个溶液中加入等体积的2.0 mol,LNaOH溶液,溶液的pH是多少?
,1解:(1) 0.20 mol,LHCl溶液的pH=0.70,要使pH = 4.0,应加入碱NaAc;
,1,1(2)加入等体积的2.0 mol,L NaAc后,生成0.10 mol,LHAc;
,1余(2.0,0.20)/2 = 0.90 mol,LNaAc;
,5pH = pK , lgc/c pH = ,lg(1.74,10) ,lg(0.10/0.90) = 5.71 aab,1,1(3)加入2.0 mol,L的HAc后, c(HAc) =1.0 mol,L
+, HAc H + Ac
1.0 ,x 0.10 +x x
,5,4,1 1.74,10 = (0.10 +x)x/(1.0,x) x =1.74,10 mol,L
+,1,4,1,1c(H) = 0.10 mol,L+1.74,10 mol,L = 0.10 mol,L pH = 0.10
,1(4)反应剩余NaOH浓度为0.9 mol,L
pOH = ,lg0.9 = 0.05 pH = 14.00,0.05 = 13.95
,1,118.0.5000 mol,L HNO溶液滴定0.5000mol,L NH,HO溶液。试计算滴定分数为0.50332
及1.00时溶液的pH值。应选用何种指示剂?
+解:滴定分数为0.50时,NH,HO溶液被中和一半,为NH,HO和NH的混合溶液; 32324
pOH = pK , lgc/c其中c= c bba a b,5 pOH = ,lg(1.74,10) = 4.76 pH = 14.00 , 4.75 = 9.24
,1+滴定分数为1.00时,NH,HO刚好完全被中和,溶液为0.2500 mol,L NH; 324,14,5 ,10 +K(NH)= K/K= 1.00,10/1.74,10= 5.75,10cK>20K c/K>500 a4wb aawaa
,,10,5 pH= 4.92 c(H),cK,0.2500,5.75,10,1.20,10a
可选指示剂:甲基红较好(4.4~6.2);溴甲酚绿(3.8~5.4)。
-19.人体中的CO在血液中以HCO和HCO 存在,若血液的pH为7.4,求血液中 HCO223323-,与HCO 的摩尔分数x(HCO)、x(HCO)? 3233
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第四章习题
,3,11解:HCO的K= 4.2,10 (pK= 6.38); K= 5.6,10 (pK= 10.25) 23a1 a1 a2 a2
,pH = pK , lgc/c 7.4 = 6.38 , lg c(HCO)/c(HCO) aab233
(HCO)(HCO)cn2323, n(HCO) = 0.095n(HCO) ,,0.095233,,(HCO)(HCO)cn33
(HCO)0.095(HCO)nn,(HCO) ,,,0.087x23323,,,(HCO),(HCO)0.095(HCO),(HCO)nnnn23333 ,,(HCO)(HCO)nn,33(HCO),,,0.91x3,,,(HCO),(HCO)0.095(HCO),(HCO)nnnn23333
,或 x(HCO) = 1, x(HCO) = 1, 0.087 = 0.913 323
20.回答下列问题并说明理由。
(1)将NaHCO 加热至270~300?,以制备NaCO基准物质,如果温度超过300?,323
部分NaCO分解为NaO,用此基准物质标定HCl溶液,对标定结果有否影响?232
为什么?
(2)以HCO,2HO来标定NaOH浓度时,如草酸已失去部份结晶水,则标定所得NaOH2242
的浓度偏高还是偏低?为什么?
(3)NHCl或NaAc含量能否分别用碱或酸的标准溶液来直接滴定? 42,(4)NaOH标准溶液内含有CO,如果标定浓度时用酚酞作指示剂,在标定以后测定3
物质成份含量时用甲基橙作批示剂,讨论其影响情况及测定结果误差的正负。
解:(1)若NaCO部分分解为NaO,由于失去了CO,则消耗比NaCO时更多的HCl,232223
即所用HCl的体积增加,使HCl的浓度偏低。
(2)HCO,2HO失去部分结晶水,则相同质量的草酸消耗的NaOH体积增加,使NaOH2242
的浓度偏低。
,8(3)由于NH,HO可以直接滴定,故NHCl不可直接滴定(cK<10); 324aa,8同理NaAc的cK<10,不可直接滴定。 ab2,(4)若NaOH溶液内含CO,标定时用酚酞作指示剂,而测定时用甲基橙为指示剂。32,由于标定时CO在NaOH中以NaCO存在, NaCONaHCO ,323233
2NaOH + CO NaCO+ HO 2molNaOH~1molNaCO~1mol HCl ,2 23 223
即使V(NaOH)值上升,标定计算出的c(NaOH)下降,(以NaOH与HCl反应为例,
别的物质也一样)。
测定时用甲基橙为指示剂,NaCO+ 2HCl2NaCl+ HO +CO,23 22 1molNaCO~2mol HCl~2molNaOH 即CO不构成影响测定结果,由于与c成正232
比,c下降故结果产生负误差。
21.下列酸或碱能否准确进行滴定?
,1,1(1)0.1 mol,LHF; (2)0.1 mol,LHCN;
,1,1(3) 0.1 mol,L NHCl; (4) 0.1 mol,L CHN(吡啶); 455,1(5) 0.1 mol,L NaAc;
,4,1,5,8 解:(1) K(HF)= 6.6,10 c = 0.1 mol,L cK = 6.6,10>10故可准确进行滴定; aaaa,10 ,1 ,11,8 (2) K(HCN)=6.2,10c = 0.1 mol,LcK = 6.2,10<10 故不可准确进行滴定; aaa
4-5
无机及分析化学学习指导
,5 ,14,5 ,10 ,1 (3) K(NH)=1.74,10K =K/K=1.00,10/1.74,10= 5.75,10c= 0.1 mol,Lb3awba,11,8 cK=5.75,10<10 故不可准确进行滴定; aa ,9 ,1,10,8(4) 吡啶K=1.7,10c=0.1mol,L cK =1.7,10<10 故不可准确进行滴定; bbbb,5 ,14,5 ,10 ,1(5) K(HAc)=1.74,10K= 1.00,10/1.74,10= 5.75,10c= 0.1mol,L ab b,11,8 cK= 5.75,10<10故不可准确进行滴定 。bb
22.下列多元酸或混合酸的溶液能否被准确进行分步滴定或分别滴定?
,1,1(1)0.1mol,L HCO; (2)0.1 mol,L HS; 2242,1,1(3)0.1 mol,L 柠檬酸; (4)0.1 mol,L 酒石酸;
,1,1,1,1(5)0.1 mol,L 氯乙酸+0.1 mol,L 乙酸;(6) 0.1 mol,L HSO+0.1 mol,L HBO; 2433,2,5,1解:(1) HCO K=5.9,10 K=6.4,10 c= 0.1 mol,L 224a1a2a,8,8,54cK>10 cK>10 K/K=5.9,10,2/6.4,10<10 故不可准确进行分步滴定; a1aa2a1a2,7,13 ,1(2) HS K=1.07,10 K=1.3,10c = 0.1 mol,L 2a1a2a,8 ,8cK>10 cK<10 故不可准确进行分步滴定,但可滴定至HS; aa1aa2,4,5 ,1,8 ,8(3) 柠檬酸 K=7.4,10 K=1.7,10c = 0.1 mol,L cK>10 cK>10 a1a2aaa1aa2,4,54 但K/K=7.4,10/1.7,10<10故不可准确进行分步滴定; a1a2 ,4 ,5 ,1,8 ,8(4) 酒石酸K=9.1,10 K=4.3,10c = 0.1 mol,L cK>10 cK>10 a1a2a1a2,4,5 4但 K/K= 9.1,10/4.3,10<10 故不可准确进行分步滴定; a1a2 ,3,5 (5) CHCOOCl的K,=1.4,10 CHCOOH的 K,,=1.74,103a3a,3,5 4K,/K,, = 1.4,10/1.8,10<10 故不可准确进行分别滴定; aa,2 (6)HSO的第一级离解可认为完全离解 K=1.2,10 24 a2,10HBO的K=5.8,10 K、K很小,忽略。HSO的二级离解常数很大,故产33a1a2a324
生一个终点,HBO不干扰。而HBO的K很小,故不可直接滴定。 3333a
23.已知某试样可能含有NaPO,NaHPO和惰性物质。称取该试样1.0000g,用水溶解。3424,1试样溶液以甲基橙作指示剂,用0.2500mol,L HCl溶液滴定,用去了32.00mL。含同
样质量的试样溶液以百里酚酞作指示剂,需上述HCl溶液12.00mL。求试样中NaPO34和NaHPO的质量分数。 24
解:由题意可得:试样中含NaPO和NaHPO(不可能含NaHPO) 342424
NaPO+ HCl = NaHPO+ NaCl (百里酚酞) 34 24
NaHPO+ HCl = NaHPO + NaCl (甲基橙) 24 24,3,1,1w(NaPO) = 12.00,10L,0.2500 mol,L,163.94g,mol/1.0000g = 0.4918 34,3,1,1w(NaHPO) = (32.00,12.00,2),10L,0.2500 mol,L,141.96g,mol/1.0000g = 0.2839 24
24.称取2.000g干肉片试样,用浓HSO煮解(以汞为催化剂)直至其中的氮素完全转化为24
硫酸氢铵。用过量NaOH处理,放出的NH吸收于50.00mL HSO(1.00mL相当于324
0.01860gNaO)中。过量酸需要28.80mL的NaOH(1.00mL相当于0.1266g邻苯二甲酸2
氢钾)返滴定。试计算肉片中蛋白质的质量分数。(N的质量分数乘以因数6.25得蛋白质的质量分数)。
,1,4解:0.01860g NaO的物质的量n(NaO) = 0.01860g/61.98g,mol = 3.001,10mol 22,1,40.1266g 邻苯二甲酸氢钾的物质的量n(邻) = 0.1266g/204.22g,mol = 6.199,10mol
,4,1 ,250.00mL HSO的物质的量n(HSO) = 50.00mL,3.001,10mol,mL= 1.501,10mol 2424,4,1,228.80mLNaOH的物质的量n(NaOH) = 28.80mL,6.199,10mol,mL= 1.785,10mol
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第四章习题
+1mol HSO~2molNaOH N~NH~1/2HSO~NaOH 24424
n(NaOH) = 2n(HSO) ; n(NH)=2n(HSO) 243 24,2,2w(蛋白质)= 6.25, [14.007, (1.501,10mol ,1.785,10 mol/2),2]/2.000 = 0.5327 25.有一在空气中暴露过的氢氧化钾,经分析测定内含水7.62%,KCO 2.38%和23.,1KOH90.00%。将此样品1.000g加1.000 molLHCl溶液46.00mL,过量的酸再用1.070
,1mol,LKOH溶液回滴至中性。然后将此溶液蒸干,问可得残渣多少克?
解:1.000g样品中m(KCO) = 0.0238g;m(KOH) = 0.90g; 23,1则:n(KCO)= 0.02380g/138.21g,mol = 0.0001722mol 23 ,1n(KOH)= 0.9000g/56.106 g,mol = 0.01604mol
KCO+ 2HCl = 2KCl + HO + CO 2n(KCO) = n(HCl) n(HCl) = n(KOH) 23 2223
故KCO与KOH共同消耗HCl n(HCl)= 2, 0.0001722 mol +0.01604 mol =0.01638mol 23,3.,1HCl共46.00,10L,1.000molL=0.04600mol;
剩余HCl共0.04600mol, 0.01638mol = 0.02962mol;
HCl与KCO 及KOH作用生成KCl 0.01638mol ; 23
过量的HCl用KOH回滴生成KCl0.02962mol;
所以生成KCl的物质的量为n(KCl) = 0.04600mol
,1 m(KCl) = 0.04600mol , 74.55g,mol= 3.429g 即可得残渣3.429g。
,126.称取混合碱试样0.8983g,加酚酞指示剂,用0.2896 mol,LHCl溶液滴定至终点,计耗去酸溶液31.45mL。再加甲基橙指示剂,滴定至终点,又耗去24.10mL酸。求试样中各组分的质量分数。
解:分析:用酚酞耗去31.45mL盐酸,再用甲基橙耗去24.10盐酸,可知试样有NaOH和
NaCO组成。 23,3w(NaCO) = 24.10,10,0.2896,105.99/0.8983 = 0.8235 23,3w (NaOH)=[(31.45,24.10),10,0.2896,40.01]/0.8983 = 0.09481
,,,27.有一三元酸,其pK= 2.0,pK= 6.0,pK=12.0。用氢氧化钠溶液滴定时,第一a1a2a3
和第二化学计量点的pH分别为多少?两个化学计量点附近有无pH突跃?可选用什么指示剂?能否直接滴定至酸的质子全部被作用?
,2 ,6,12,,,解:K=10 K=10 K=10 a1a2a3
+,2,6,4,,第一计量点时 c(H)= pH= 4 KK,10,10,10a1a2
+,6,12,9,,第二计量点时 c(H)= pH= 9 K K ,10,10,10a2a3
,8,7 ,8,1,,由于cK>10 cK=10>10 (含c=0.1mol,L) a1a2
故两个计量点产生pH突跃,但第二计量点的突跃可能较小,最好用混合指示剂,
第一计量点用甲基橙,第二计量点时用百里酚酞+酚酞。
,13 ,8,由于cK=10<<10,因而不能滴定至酸的质子全部被作用。 a3
28.某一元弱酸(HA)试样1.250g,用水溶解后定容至50.00mL,用41.20mL0.0900
,1mol,LNaOH标准溶液滴定至化学计量点。加入8.24mLNaOH溶液时,溶液pH为4.30。求:
,1(1)弱酸的摩尔质量(g,mol) (2)弱酸的解离常数
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无机及分析化学学习指导
(3)化学计量点的pH值 (4)选用何种指示剂 解:(1) HA + NaOH = NaA + HO 2,3,11.250g/M = 41.20,10L,0.0900 mol,L
,3,1M = 1.250g/[41.20,10,0.0900]mol =337g,mol
,(2) pH = pK, lgc/c a ab,pK= pH + lgc/ca ab ,3,3= 4.30 + lg[(41.20,8.24),10,0.0900]/[8.24,10,0.0900]
= 4.30 + lg(32.96/8.24)
= 4.90
,4.90 ,5,K= 10= 1.3,10 a ,14.00,4.90,9.10,,,(3) K= K/K= 10/10=10 bwa
,9.10,6,0.0900,41.20计量点时c(OH)= cK,,10,5.68,10b50.00,41.20
,6 pH = 14.00 , pOH =14.00 +lg 5.68,10= 8.75
(4) 可选酚酞为指示剂。
,129.What is the pH at 25? of a solution which is 1.5 mol,L with respect to formic acid and 1.0
,1,mol,Lwith respect to sodium formate? pK for formic acid is 3.751 at 25?。 a,1,1,解:已知甲酸的pK = 3.751,求1.5mol,L甲酸的pH值和1mol,L甲酸钠的pH值。 a,1,,甲酸的c=1.5 mol,L cK>20K c/K>500 aaa waa
+,3.751c(H)= cK,1.5,10,0.0163aa
pH=1.79
,14.00,3.751,10.25,,,甲酸钠的K= K/K= 10/10=10 bwa ,1,,,c=1.0mol,L cK>20K c/K>500 b bbwbb
,,10.25,6c(OH)= cK,1.0,10,7.5,10bb
pOH = 5.13 pH = 14.00,5.2 = 8.87 30.Calculate the concentration of sodium acetate needed to produce a pH of 5.0 in a solution of
,1acetic acid (0.10mol,L) at 25?. pK for acetic acid is 4.756 at 25?. a
解: pH = pK, lgc/c a ab
lgc/c = pK, pH = 4.756 , 5.0 = ,0.24 c/c = 0.57 abaab,1c = 0.10/0.57 = 0.18(mol,L) b,131.Calculate the percent ionization in a 0.20 mol,L solution of hydrofluoric acid, HF
,4(K=7.2×10)。 a
22,c,4 0.20,a解: K= 7.2,10= a1,,1,,
2,4,4 0.20, + 7.2,10, , 7.2,10= 0
, = 0.058
32.The concentration of HS in a saturated aqueous solution at room temperature is 2,1+,2,approximately 0.10 mol,L. Calculated c(HO),c(HS),and c(S) in the solution. 3
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第四章习题
,7,13 解:HS的K=1.1,10 K=1.3,10 2a1a2,8 cK=1.3,10>20Kc/K>500 aa1w aa1
+,7,4,1c(H)=(mol,L) cK,0.10,1.1,10,1.0,10aa1
,+ 2, HS H+ S
,4,41.0,10,x 1.0,10+x x
,4x(1.0, 10,x),13,13 x =1.3,10 ,1.3,10,41.0 ,10,x2,,13,1即c(S) = 1.3,10mol,L
,,4 ,13 ,4 ,1c(HS) = 1.0,10, 1.3,10= 1.0,10mol,L 33.Calculate the equilibrium concentration of sulfide ion in a saturated solution of hydrogen
sulfide to which enough hydrochloric acid has been added to make the hydronium ion
,1concentration of the solution 0.1 mol,L at equilibrium. ( A saturated HS solution is 0.10 2,1mol,L in hydrogen sulfide ).
+, ,7解: (1) HS H + HSK=1.1,10 2a1 ,+2,,13(2) HS H + S K=1.3,10 a2+2,(1) + (2) HS 2H + S 2 2,2,c(H)c(S),7,13,20K,K,K,,1.1,10,1.3,10,1.4,10 a1a2c(HS)2
c(HS),K22,,20,19,10.10 c(S),,1.4,10,,1.4,10(mol,L)2,2c(H)0.1034.Calculate the hydroxide ion concentration, the percent reaction, and the pH of a 0.050
,1,5mol,L solution of sodium acetate. (For acetic acid, K=1.74×10) . a,14.00,5 ,10解:NaAc的K=10/1.74,10= 5.75,10 b,10,11,10 cK = 0.050,5.75,10 = 2.88,10>20Kc/K= 0.050/5.75,10>500 bbw bb
,10,6,c(OH0.050,5.75,10,5.4,10)=
pOH=5.27 pH = 14.00,5.27=8.73
+,9,1c(H) = 1.9,10(mol,L)
,,6,4the percent reaction , = c(OH)/c= 5.4,10/0.050 = 1.1,10 豆丁致力于构建全球领先的b
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